# Exact Pole-Zero Cancellation in a SISO Transfer Function

Pole-Zero cancellation is one of those phantoms in control which are often either sidestepped or only treated informally. Personally, I've never deeply understood the reasoning behind some of the rules--in fact, there isn't even really a consensus on what that reasoning is.

For the purpose of this question, suppose $G$ is a SISO transfer function. Let $p$ be a pole which is exactly cancelled by a zero at the same location. We treat the system $G$ formally in the sense that we assume 100% fidelity to whatever we are trying to model. If you prefer to think of the system in terms of differential equations, this is equivalent to saying the differential equations perfectly model the system. The question is whether or not the system $\tilde{G}$ obtained from cancelling $p$ is distinguishable from $G$.

My answer is no by the following reasoning.

If $G$ has 100% fidelity then every system parameter is uniquely determined by $G$. It follows that if $H$ is another transfer function and $H(s) = G(s)$ for all $s$, then $H$ is mathematically indistinguishable from $G$ and by extension so too are their system parameters.

As regard $G$ and $\tilde{G}$ we have $$G(s) = \frac{s-p}{s-p}\tilde{G}(s)$$ by their definition. Clearly $G(s) = \tilde{G}(s)$ whenever $s\neq p$, so we need only check that $G(p) = \tilde{G}(p).$ To do so we may evaluate $$\lim_{s\rightarrow p}G(s) = \lim_{s\rightarrow p}\frac{(s-p)\tilde{G}(s)}{s-p},$$ which by L'Hopital's rule is equivalent to $$\lim_{s\rightarrow p}\ (s-p)\tilde{G}'(s) + \lim_{s\rightarrow p}\ \tilde{G}(s) = \tilde{G}(p).$$ This calculation finds $G = \tilde{G}$, showing the system before and after pole-zero cancellation are equivalent.

I therefore conclude that exact SISO pole-zero cancellation under the assumption of 100% model fidelity has no affect on the system.

What are the thoughts of others?

Edit

Stelios has made some thoughtful points regarding the nature of $p$ as a removable singularity. Unfortunately some of the conventions I've used also appear to have caused confusion, and I will clear them up presently.

(1) When I write $G(p)$ I mean $\lim_{s\rightarrow p}G(s)$. I concede that it is unclear whether showing $\lim_{s\rightarrow p}G(s) = \lim_{s\rightarrow p}\tilde{G}(s)$ is equivalent to saying $G(s) = \tilde{G}(s)$ for all $s$, since, as Stelios has argued, it is unclear whether or not $G(p)$ can be assigned any meaning in this way.

(2) Supposing that two transfer functions $G$ and $H$ can be assigned meaning for every $s$ in a domain which they share, it is no more than a basic set theoretic statement to define $G =H$ iff $G(s) = H(s)$ for all $s$. In fact, if $A$ and $B$ are sets and $f$ maps $A$ to $B$, then WLOG we may regard $f$ as a subset of $A \times B$ consisting of the points $(x,f(x))$ for $x \in D(f)$. If $g$ is another function and $D(g) = D(f)$, then if $g(x) = f(x)$ for each $x$ in their domain it follows that $f$ and $g$ describe the same subset of $A \times B$. This is surely enough to say $f = g$.

I presume you consider rational transfer functions as typically employed in signal processing and control theory.

I think your reasoning is invalid due to the following conceptual error. You "prove" that $\tilde{G}(s)$ and $G(s)$ are "mathematically equivalent" (which, according to your definition, means that $\tilde{G}(s)=G(s)$ for all $s$), yet $G(s)$ is not formally defined at $s=p$. Therefore, a comparison between $G(s)$ and $\tilde{G}(s)$ at $s=p$ makes no sense.

What you are essentially asking is whether a function $G$, that is analytic in a domain $\Omega$ except a single (isolated) point $s_0$, can be "replaced" by another function $F$ that is exactly the same over $\Omega\setminus \{s_0\}$, yet appropriately defined at $s_0$ such that it is analytic over all $\Omega$ (including $s_0$). Why we want to do so? Because it makes life easier and, for the majority of applications, this replacement has not effect. For a trivial example, why would one use the function $f(s)=s/s$, defined everywhere expect $s=0$, instead of $f(s)=1$, defined for all $s$?

The above question is a standard one in (complex) analysis and is related to the concepts of singular points (and Laurent series). The theory says that in the case of $s_0$ being a removable singularity for $G$, we can replace $G$ by $F$, defined as $F(s)=G(s)$ for all $s\neq s_0$ and $F(s_0)\triangleq\lim_{s\rightarrow s_0}G(s)$.

In your example, $p$ is indeed a removable singularity with $\lim_{s\rightarrow s_0}G(s)=\tilde{G}(s_0)$, therefore we can "replace" $G(s)$ by $\tilde{G}(s)$ which is equal to $G(s)$ for all $s\neq s_0$, yet also analytic at $s=s_0$.

• To address the "conceptual error", I am associating $G(p)$ with $\lim_{s\rightarrow p}G(s)$. I understand the point however. In this case the issue is whether or not a system whose Laplace transform possesses a removable singularity is equivalent to the system obtained from the Laplace inversion of a holomorphic extension of the former system over the singularity. Perhaps that is how the issue of "hidden modes" could be better addressed. – Mortified Through Math Apr 5 '17 at 21:43

If you want to be rigorous on this you have to distinguish between a system, say $\Sigma$, and its representation, say $G$ as transfer function or $(A,B,C,D)$ in state-space. Of course, the system you are suggesting (the one with transfer function $\tilde{G}$), say $\tilde{\Sigma}$, will have the same transfer function $\tilde{G}=G$ since the pole gets cancelled. However, the systems are not the same since $\Sigma$ contains the unobservable and/or uncontrollable mode $p$. They only exhibit the same input-output-behaviour.

• I disagree. If you agree that a set of ODEs can rigorously describe a system, you also agree that the transfer function obtained from this set can describe the system (and the reverse) since the Laplace transform is bijective. If we assume the idealized case where the ODEs have 100% fidelity, then it is pointless to distinguish between $\Sigma$ and its describing ODEs. In the real world there is obviously a difference, but the point of this question is to determine the purely theoretical implications of pole-zero cancellation. – Mortified Through Math Apr 5 '17 at 23:25
• As a map from a state-space-model to a transfer function the Laplace transform is not bijective in the following sense: starting from a state-space representation, when computing the transfer function $C(sI-A)^{-1}B+D$, pole-zero-cancellations take place if and only if there are uncontrollable and/or unobservable modes. Hence, starting with a state-space description of $\Sigma$, applying the Laplace transform und then its inverse leads to an ODE-system with dimension n-1 (if $\Sigma$ had n) due to the cancellation. That's why ODEs and transfer functions are not equivalent representations. – Nukular Apr 5 '17 at 23:38
• Correct for the MIMO case but I am addressing the SISO case. Here the only actual "state" is the output, which is an element of some $L_p$ space, as is the input, and the problem is the solution to the equation $Dy = Eu$ where $D$ and $E$ are linear mappings of $L_p$ into itself. The invertibility of the Laplace transform is enough to guarantee its bijectivity on $L_p$, but since $D$ and $E$ are also linear, the Laplace transform takes the unique elements in the range $D,E$ to unique elements in the Laplace domain. In this case the failure of bijectivity using a state-representation (cont). – Mortified Through Math Apr 5 '17 at 23:55
• Is a failure of the state-representation itself to uniquely realize a given transfer function. – Mortified Through Math Apr 5 '17 at 23:55
• Why should this be wrong for the SISO case? SISO systems are a special case of MIMO system and not vice versa. Also, the number of states in the ODE system for $/Sigma$, say n again, depends on the number of poles in G. The number of outputs (1 in our case) says nothing about n... of course, as a mapping on signals, the Laplace transform is invertible. However, the failure of bijectivity in that case is not because of non-uniqueness of state-space representation since $(A,B,C,D)$ has the same transfer function than any transformed state-space system $(T^{-1}AT,T^{-1}B,CT,D)$. – Nukular Apr 6 '17 at 9:24